A straight line parallel to BC of ABC intersects AB and AC at points P and Q 53 54 respectively. AP = QC, PB= 4 units and AQ = 9 units, then the length of AP is
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As far as i know this problem is based on a theoram that is that a line parallel to the third side of a triangle cuts the other sides in a equal ratio in other words ap/pb=aq/qc.
Since ap=qc therefore we suppose it x.
Now
X/4=9/x
X*X=36
X=6
Since ap=qc therefore we suppose it x.
Now
X/4=9/x
X*X=36
X=6
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